Question

Ayuda!! Con límites

Answer 1

1) (la 2 es lo  mismo)

$L=\lim\limits_{x\to \frac{\pi}{2}}\left(\dfrac{\pi}{2}-x\right)\tan x\\ \\ \\ L=\lim\limits_{x- \frac{\pi}{2}\to 0}\left(\dfrac{\pi}{2}-x\right)\tan x\\ \\ \\ L=\lim\limits_{\xi\to 0}-\xi\tan \left(\xi+\dfrac{\pi}{2}\right)\\ \\ \\ L=\lim\limits_{\xi\to 0}\xi\cot \xi\\ \\ \\ L=\lim\limits_{\xi\to 0}\dfrac{\xi}{\tan\xi}\\ \\ \\ \boxed{L=1}$

(3)
$L=\lim\limits_{x\to 4}\dfrac{\sqrt{1+2x}-3}{x-4}\\ \\ \\ L=\lim\limits_{x\to 4}\dfrac{\sqrt{1+2x}-3}{x-4}\cdot \dfrac{\sqrt{1+2x}+3}{\sqrt{1+2x}+3}\\ \\ \\ L=\lim\limits_{x\to 4}\dfrac{\sqrt{1+2x}^2-3^2}{(x-4)\sqrt{1+2x}+3}\\ \\ \\ L=\lim\limits_{x\to 4}\dfrac{(1+2x)-9}{(x-4)\sqrt{1+2x}+3}\\ \\ \\ L=\lim\limits_{x\to 4}\dfrac{2x-8}{(x-4)\sqrt{1+2x}+3}\\ \\ \\ L=\lim\limits_{x\to 4}\dfrac{2(x-4)}{(x-4)\sqrt{1+2x}+3}\\ \\ \\ L=\lim\limits_{x\to 4}\dfrac{2}{\sqrt{1+2x}+3}\\ \\ \\ \boxed{L=\dfrac{1}{3}}$